Question

Find the equation of the straight line passing through the point (-3,2)and making an angle of 45° with the straight line 3x-y+4=0

Answer 1

tan theta = 45° so it's 1 i.e slope is 1

y - y1 = m(x - x1)

y - 2 = 1(x + 3)

y - 2 = x + 3

y - x = 3 + 2

y - x = 5

hence the equation for the line is y - x = 5

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Answer 2

The equation of the straight line passing through the point (-3,2) and making an angle of 45° with the straight line 3x - y + 4 = 0 is y = (4 - 3√2)x + (10 + 6√2).

We know that

The given line has an angle of 45 degrees with the line 3x - y + 4 = 0.

The slope of the given line can be found by finding the slope of the line 3x - y + 4 = 0 and then using the formula for the slope of a line that makes an angle θ with the x-axis:

slope of 3x - y + 4 = 0:

3x - y + 4 = 0

y = 3x + 4

slope = 3

slope of line with angle 45 degrees:

tan 45 = (slope of line - slope of 3x - y + 4 = 0) / (1 + slope of line * slope of 3x - y + 4 = 0)

1 = (slope of line - 3) / (1 + slope of line * 3)

slope of line = 4 - 3√2

Now that we have the slope of the line and a point it passes through (-3,2),

We can use the point-slope form of the equation of a line to find the equation:

y - y1 = m(x - x1) where (x1, y1) is the given point and m is the slope

We just found

y - 2 = (4 - 3√2)(x + 3)

Expanding this out and simplifying, we get:

y = (4 - 3√2)x + (10 + 6√2)

Therefore, the equation of the straight line passing through the point (-3,2) and making an angle of 45° with the straight line 3x - y + 4 = 0 is y = (4 - 3√2)x + (10 + 6√2).

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